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Consider a disk pack with a seek time of $4$ milliseconds and rotational speed of $10000$ rotations per minute (RPM). It has $600$ sectors per track and each sector can store $512$ bytes of data. Consider a file stored in the disk. The file contains $2000$ sectors. Assume that every sector access necessitates a seek, and the average rotational latency for accessing each sector is half of the time for one complete rotation. The total time (in milliseconds) needed to read the entire file is__________________
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76 votes

Since each sector requires a seek, 

Total time $= 2000$  (seek time $+$ avg. rotational latency $+$ data transfer time)

Since data transfer rate is not given, we can take that in $1$ rotation, all data in a track is read. i.e., in $60/10000 = 6\;\text{ms}, 600 \times 512$ bytes are read. So, time to read $512\;\text{bytes} = 6/600\;\text{ms} = 0.01\;\text{ms}$

$= 2000 \times (4 \; \text{ms} + 60 \times 1000 /2 \times 10000 + 0.01) $

$= 2000 \times (7.01 \ \text{ms})$

$= 14020 \ \text{ms}. $

http://www.csee.umbc.edu/~olano/611s06/storage-io.pdf

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41 votes
Seek time (given) = 4ms

RPM = 10000 rotation in 1 min [60 sec]

So, 1 rotation will be =60/10000 =6ms [rotation speed]

Rotation latency= 1/2 * 6ms=3ms

# To access a file, total time includes =seek time + rot. latency +transfer time

TO calc. transfer time, find transfer rate Transfer rate = bytes on track /rotation speed

so, transfer rate = 600*512/6ms =51200 B/ms

transfer time= total bytes to be transferred/ transfer rate

so, Transfer time =2000*512/51200 = 20ms

Given as each sector requires seek tim + rot. latency = 4ms+3ms =7ms

Total 2000 sector takes = 2000*7 ms =14000 ms To read entire file ,total time = 14000 + 20(transfer time) = 14020 ms
2 votes
2 votes

Seek time = 4ms
60s        →   10000 rotations
Rotation Time :-            (60 / 10000) = 6 ms    (in 1- rotation)
Rotational latency  :-     (1 / 2 )6ms = 3ms
1track    →   600 sectors

6ms←600 sectors (1 rotation means 600 sectors (or) 1 t
2000 sector         →   (2000) 0.01 =  20 ms
∴total time needed to read the entire file is:
= 2000 (4+3) +20
=8000+6000+20 = 14020

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